1058 Lexicographically Smallest Equivalent String

Lexicographically Smallest Equivalent String

You are given two strings of the same length s1 and s2 and a string baseStr.

We say s1[i] and s2[i] are equivalent characters.

For example, if s1 = "abc" and s2 = "cde", then we have 'a' == 'c', 'b' == 'd', and 'c' == 'e'.

Equivalent characters follow the usual rules of any equivalence relation:

**Reflexivity:** 'a' == 'a'.
**Symmetry:** 'a' == 'b' implies 'b' == 'a'.
**Transitivity:** 'a' == 'b' and 'b' == 'c' implies 'a' == 'c'.

For example, given the equivalency information from s1 = “abc” and s2 = “cde”, “acd” and “aab” are equivalent strings of baseStr = “eed”, and “aab” is the lexicographically smallest equivalent string of baseStr.

Return the lexicographically smallest equivalent string of *baseStr by using the equivalency information from s1 and *s2.

 

Example 1:

**Input:** s1 = "parker", s2 = "morris", baseStr = "parser"
**Output:** "makkek"
**Explanation:** Based on the equivalency information in s1 and s2, we can group their characters as [m,p], [a,o], [k,r,s], [e,i].
The characters in each group are equivalent and sorted in lexicographical order.
So the answer is "makkek".

Example 2:

**Input:** s1 = "hello", s2 = "world", baseStr = "hold"
**Output:** "hdld"
**Explanation: **Based on the equivalency information in s1 and s2, we can group their characters as [h,w], [d,e,o], [l,r].
So only the second letter 'o' in baseStr is changed to 'd', the answer is "hdld".

Example 3:

**Input:** s1 = "leetcode", s2 = "programs", baseStr = "sourcecode"
**Output:** "aauaaaaada"
**Explanation:** We group the equivalent characters in s1 and s2 as [a,o,e,r,s,c], [l,p], [g,t] and [d,m], thus all letters in baseStr except 'u' and 'd' are transformed to 'a', the answer is "aauaaaaada".

 

Constraints:

1 <= s1.length, s2.length, baseStr <= 1000
s1.length == s2.length
s1, s2, and baseStr consist of lowercase English letters.

class Solution:
    def smallestEquivalentString(self, s1: str, s2: str, baseStr: str) -> str:
        mp = {}

        for k1, k2 in zip(s1,s2):
            if k1 in mp and k2 in mp:
                t = mp[k1].union(mp[k2])
                for m in t:
                    mp[m] = t
                continue

            if k1 in mp:
                t = mp[k1]
                t.add(k2)
                mp[k2] = t
                mp[k1] = t
                continue

            if k2 in mp:
                t = mp[k2]
                t.add(k1)
                mp[k1] = t
                mp[k2] = t
                continue

            t = set()
            t.add(k1)
            t.add(k2)
            mp[k1] = t
            mp[k2] = t


        res = ""
        for k in baseStr:
            if k in mp:
                ls = list(mp[k])
                ls.sort()
                res += ls[0]
            else:
                res += k
        return res